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3 years ago

What is the midpoint of this segment??

Mathematics

2 answers:

bulgar [2K]3 years ago

7 0

The right answer is (6,3)

please see the attached picture for full solution

Hope it helps

Good luck on your assignment

Korvikt [17]3 years ago

3 0

Answer:

A) 6,3

Why?

The midpoint, which is somewhat self-explanatory thanks to its name, is a point in the center of a line.

To answer this question properly, you need to analyze the line and assess the situation.

How did it go from x = 2 to x = 10? Let's see, if we count by two's across the line, you'll notice it makes more sense. So now we know that every square goes up by two. x = 6 is the exact center of the line.

I hope this wasn't confusing, it's tough to explain it!

You might be interested in

The table below shows the numbers of tickets sold at a movie theater on Friday.

sleet_krkn [62]

Answer:

Number of Adult's tickets sold on Saturday = 3,356

Number of Children's tickets sold on Saturday = 2, 928

Total number of tickets sold over these two days is 8,938.

Step-by-step explanation:

Here, the number of tickets sold on FRIDAY:

Adult Ticket sold = 1,678

Children's Tickets sold = 976

So, the total number of tickets sold on Friday

= Sum of ( Adult + Children's ) tickets = 1,678 + 976 = 2,654 .... (1)

The number of tickets sold on SATURDAY:

Adult Ticket sold = 2 times as many as the number of adult tickets sold

on Friday

= 1,678 x 2 = 3,356

Children's Tickets sold = 3 times as many as the number of children's

tickets sold on Friday.

= 976 x 3 = 2, 928

So, the total number of tickets sold on Saturday

= Sum of ( Adult + Children's ) tickets = 3,356 + 2,928 = 6, 284 .... (2)

Now, the total number of tickets booked in these two days :

Sum of tickets booked on (Friday + Saturday)

= 2,654 + 6, 284 = 8,938

Hence, total number of tickets sold over these two days is 8,938.

7 0

3 years ago

PLEASE HELP!!!

andrew-mc [135]

Answer:

(-5.25,-3.5)

Step-by-step explanation:

Q(-3,-2)

Q'(-3*7/4.-2*7/4)

4 0

3 years ago

The question is what is other equation to find 16-9

disa [49]

Its 9-1 so that would be 8 idk if you are right just noticed and though it would help

4 0

3 years ago

Use this list of Basic Taylor Series and the identity sin2θ= 1 2 (1−cos(2θ)) to find the Taylor Series for f(x) = sin2(3x) based

notsponge [240]

Answer:

The Taylor series for $sin^2(3 x) = - \sum_{n=1}^{\infty} \frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$ , the first three non-zero terms are $9x^{2} -27x^{4}+\frac{162}{5}x^{6}$ and the interval of convergence is $( -\infty, \infty )$

Step-by-step explanation:

These are the steps to find the Taylor series for the function $sin^2(3 x)$

Use the trigonometric identity:

$sin^{2}(x)=\frac{1}{2}*(1-cos(2x))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(2(3x)))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$

2. The Taylor series of $cos(x)$

$cos(y) = \sum_{n=0}^{\infty}\frac{-1^{n}y^{2n}}{(2n)!}$

Substituting y=6x we have:

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

3. Find the Taylor series for $sin^2(3x)$

$sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$ (1)

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$ (2)

Substituting (2) in (1) we have:

$\frac{1}{2} (1-\sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!})\\ \frac{1}{2}-\frac{1}{2} \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

Bring the factor $\frac{1}{2}$ inside the sum

$\frac{6^{2n}}{2}=9^{n}2^{2n-1} \\ (-1^{n})(9^{n})=(-9^{n} )$

$\frac{1}{2}-\sum_{n=0}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

Extract the term for n=0 from the sum:

$\frac{1}{2}-\sum_{n=0}^{0}\frac{-9^{0}2^{2*0-1}x^{2*0}}{(2*0)!}-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \frac{1}{2} -\frac{1}{2} -\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ 0-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ sin^{2}(3x)=-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

To find the first three non-zero terms you need to replace n=3 into the sum

$sin^{2}(3x)=\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \sum_{n=1}^{3}\frac{-9^{3}2^{2*3-1}x^{2*3}}{(2*3)!} = 9x^{2} -27x^{4}+\frac{162}{5}x^{6}$